What is the empirical formula of a compound containing #C, H#, and #O# if combustion of 3.69 g of the compound yields 5.40 g of #CO_2# and 2.22 g of #H_2O#?

Answer 1

We get finally an empirical formula of #CH_2O#; I think the question is suspect.

ONLY the #C# and the #H# of the combustion can be presumed to derive from the unknown. (Why? Because the analysis is performed in air and typically an oxidant is added to the combustion.)
#"Moles (i) and mass (ii) of carbon"# #=# #(5.40*g)/(44.01*g*mol^-1)# #=# #0.123*mol# #-=# #1.47*g*C#
#"Moles (i) and mass (ii) of hydrogen"# #=# #2xx(2.22*g)/(18.01*g*mol^-1)# #=# #0.247*mol# #-=# #0.249*g*H#. Note that the hydrogen in the compound was combusted to water; this is why we multiply the molar quantity by 2.
And now, finally, we work out the percentage composition of #C,H,O# with respect to the original sample:
#%C# #=# #(1.47*g)/(3.69*g)xx100%=40.00%#
#%H# #=# #(0.249*g)/(3.69*g)xx100%=6.75%#
#%O# #=# #(100-39.84-6.75)%=53.25%#
Note that we cannot (usually) measure the percentage of oxygen in a microanalysis as extra oxidant is typically added. Thus #O%# is the percentage balance.
After all this wrok we start again. We assume that there were #100*g# of compound. And from this we work out the empirical formula.
#"Moles of carbon"# #=# #(40.0*g)/(12.011*g*mol^-1)# #=# #3.33*mol*C#.
#"Moles of hydrogen"# #=# #(6.75*g)/(1.00794*g*mol^-1)# #=# #6.70*mol*H#.
#"Moles of oxygen"# #=# #(53.41*g)/(15.999*g*mol^-1)# #=# #3.34*mol*H#.
If we divide thru by the smallest molar quantity, we get, an empirical formula of #CH_2O#. I am not terribly satisifed with this question. A molecular mass should have been quoted. This is a lot of work for simple sugar.
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Answer 2

To determine the empirical formula of the compound, we need to find the moles of carbon, hydrogen, and oxygen in the compound.

Given: Mass of CO2 produced = 5.40 g Mass of H2O produced = 2.22 g

  1. Calculate the moles of carbon in CO2: Molar mass of CO2 = 12.01 g/mol (C) + 2(16.00 g/mol) (O) = 44.01 g/mol Moles of carbon = Mass of CO2 / Molar mass of CO2 = 5.40 g / 44.01 g/mol = 0.1227 mol

  2. Calculate the moles of hydrogen in H2O: Molar mass of H2O = 2(1.01 g/mol) (H) + 16.00 g/mol (O) = 18.02 g/mol Moles of hydrogen = Mass of H2O / Molar mass of H2O = 2.22 g / 18.02 g/mol = 0.1232 mol

  3. Determine the moles of oxygen: The moles of oxygen can be found by subtracting the sum of moles of carbon and hydrogen from the total moles of CO2 and H2O. Total moles of CO2 and H2O = 0.1227 mol (C) + 0.1232 mol (H) = 0.2459 mol Moles of oxygen = Total moles of CO2 and H2O - (moles of carbon + moles of hydrogen) = 0.2459 mol - (0.1227 mol + 0.1232 mol) = 0.0000 mol

  4. Determine the ratio of moles of each element to the smallest number of moles: Since the moles of oxygen are negligible, we consider the moles of carbon and hydrogen. The smallest number of moles is 0.1227 mol. So, the ratio of moles of carbon to the smallest number of moles is 1, and the ratio of moles of hydrogen to the smallest number of moles is approximately 1.

  5. Write the empirical formula: The empirical formula is CH.

Therefore, the empirical formula of the compound containing C, H, and O is CH.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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